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Physics 1202: Lecture 21
Today’s Agenda
• Announcements:
– Lectures posted on:
www.phys.uconn.edu/~rcote/
– HW assignments, etc.
• Homework #6:
– Due next Friday
A prism does two
things,
1. Bends light the
same way at both
entrance and exit
interfaces.
2. Splits colors due to
dispersion.
Index of refraction
Prisms
1.54
ultraviolet
absorption
bands
1.52
1.50
frequency
white light
prism
Prisms
Entering
q1
Exiting
q3
q2
For air/glass interface, we
use n(air)=1, n(glass)=n
q4
Prisms
Overall Deflection
f
q1
q3
q4
q2
• At both deflections the amount of downward deflection
depends on n (and the prism apex angle, f).
• The overall downward deflection goes like,
g ~ A(f) + B n
• Different colors will bend different amounts !
Lecture 21, ACT 1
White light is passed through a
prism as shown. Since n(blue)
> n(red) , which color will end
up higher on the screen ?
A) BLUE
B) RED
?
?
LIKE SO!
In second rainbow
pattern is reversed
Total Internal Reflection
– Consider light moving from glass (n1=1.5) to air (n2=1.0)
n1
incident
ray
q1 qr
reflected
ray
GLASS
q2
refracted
ray
n2
AIR
ie light is bent away from the normal.
as q1 gets bigger, q2 gets bigger, but
q2 can never get bigger than 90° !!
2
In general, if sin q1  sin qC = (n2 / n1), we have NO refracted ray;
we have TOTAL INTERNAL REFLECTION.
For example, light in water which is incident on an air surface with
angle q1 > qc = sin-1(1.0/1.5) = 41.8° will be totally reflected. This
property is the basis for the optical fibers used in communication.
ACT 2: Critical Angle...
An optical fiber is
cladded by another
dielectric. In case I this
is water, with an index
of refraction of 1.33,
while in case II this is
air with an index of
refraction of 1.00.
Compare the critical
angles for total internal
reflection in these two
cases
a) qcI>qcII
b) qcI=qcII
c) qcI<qcII
water n =1.33
Case I
qc
glass n =1.5
water n =1.33
air n =1.00
Case II
qc
glass n =1.5
air n =1.00
ACT 3: Fiber Optics
The same two fibers are
used to transmit light
from a laser in one
Case I
room to an experiment
in another. Which
makes a better fiber,
the one in water (I) or
the one in air (II) ?
a) I Water
b) II Air
Case II
water n =1.33
qc
glass n =1.5
water n =1.33
air n =1.00
qc
glass n =1.5
air n =1.00
Problem
You have a prism that from the side forms a
triangle of sides 2cm x 2cm x 22cm, and
has an index of refraction of 1.5. It is
arranged (in air) so that one 2cm side is
parallel to the ground, and the other to the
left. You direct a laser beam into the prism
from the left. At the first interaction with the
prism surface, all of the ray is transmitted
into the prism.
a) Draw a diagram indicating what happens to
the ray at the second and third interaction
with the prism surface. Include all reflected
and transmitted rays. Indicate the relevant
angles.
b) Repeat the problem for a prism that is
arranged identically but submerged in water.
A) Prism in air
Solution
• At the first interface q=0o, no deflection of initial light direction.
• At 2nd interface q=45o, from glas to air ?
• Critical angle: sin(qc)=1.0/1.5 => qc= 41.8o < 45o
• Thus, at 2nd interface light undergoes total internal reflection
• At 3rd interface q=0o, again no deflection of the light beam
B) Prism in water (n=1.33)
• At the first interface q=0o, the same situation.
• At 2nd interface now the critical angle: sin(qc)=1.33/1.5 => qc= 62o > 45o
• Now at 2nd interface some light is refracted out the prism
•
n1 sin(q1) = n2 sin(q2)
=> at q2 = 52.9o
• Some light is still reflected, as in A) !
• At 3rd interface q=0o, the same as A)
h
q
q
R-i
h’
o-R
i
o
&
h
i
o
f
h’
Nothing New!
• For the next few lectures we will be studying geometric optics. You
should be comforted by the fact that you already know the underlying
fundamentals of what is going on.
– Namely, you know how light propagates in situations in which the length
scales are much greater than the light’s wavelength.
– Reflection:
– Refraction:
incident
ray
q1 qr
q2
refracted
ray
reflected
ray
n1
n2
• We will use these laws to understand the properties of mirrors
(perfect reflection) and lenses (perfect refraction).
• We will also discover properties of combinations of lenses which
will allow us to understand such applications as microscopes,
telescopes and eyeglasses.
1
Flat Mirror
Plane
Mirror
Virtual
Image
Object
REAL
q
q
q
q
o
i
o = -i
VIRTUAL
Flat Mirror
Images of Extended Objects
Plane
Mirror
Virtual
Image
Extended
Object
h
h’
o
i
o = -i
REAL
VIRTUAL
Magnification:
M = h’/ h = 1
Multiple Reflection
180o
Image 6
Image 4
Image 3
Image 2
6
4
2
Object
Image 5
Image 1
1
3
5
Lecture 21, ACT 4
• Let’s now consider a curved mirror.
We start with CONVEX mirror.
4A
– Where do the rays which are
reflected from the convex mirror
shown intersect?
(a) to left of
(b) to right of
R
(c) they don’t intersect
Lecture 21, ACT 4
4B • What is the nature of the image of the arrow?
(a) Inverted and in front of the mirror
(b) Inverted and in back of the mirror
(c) Upright and in back of the mirror
R
Concave Spherical Mirrors
• We start by considering the reflections from a concave spherical
mirror in the paraxial approximation (ie small angles of incidence
close to a single axis):
• First draw a ray (light blue)
from the tip of the arrow through
q
q
the center of the sphere. This
ray is reflected straight back

since the angle of incidence = 0.

• Now draw a ray (white) from
the tip of the arrow parallel to
the axis. This ray is reflected
with angle q as shown.
• Note that the two rays intersect in a point, suggesting an inverted image.
• To check this, draw another ray (green) which comes in at some angle 
that is just right for the reflected ray to be parallel to the optical axis.
• Note that this ray intersects the other two at the same point, as it
must if an image of the arrow is to be formed there.
• Note also that the green ray intersects the white ray at another point
along the axis. We will call this point the focal point ( ).
The Mirror Equation
• We will now transform the geometric drawings into algebraic
equations:
from triangles,
eliminating ,


q
g
b
object
h
image
i
o
Now we employ the small angle approximations:
Plugging these back into the above equation relating the angles, we get:
Defining the focal
length f = R/2,
This eqn is known as the mirror eqn. Note that there is no mention of q in
this equation. Therefore, this eqn works for all q, ie we have an image!
Magnification
• We have derived the mirror eqn which determines the image
distance in terms of the object distance and the focal length:
• What about the size of the image?
h
• How is h’ related to h??
• From similar triangles:
q
q
h’
o
Now, we can introduce a sign convention. We can
indicate that this image is inverted if we define its
magnification M as the negative number given by:
i
More Sign Conventions
• Consider an object distance s which is less than the focal
length:
Ray Trace:
• Ray through the center of
the sphere (light blue) is
q
reflected straight back.
h q o
• Ray parallel to axis (red)
i
passes through focal point f.
f
h’
• These rays diverge! ie
these rays look they are
coming from a point behind
the mirror.
• We call this a virtual image, meaning that no light from the object passes
through the image point.
• Proof left to student: This situation is described by the same mirror
equations as long as we take the convention that images behind the mirror
have negative image distances s’. ie:
In this case, i < 0, which leads to M > 0,
indicating that the image is virtual (i<0)
and not inverted (M>0).
Concave-Planar-Convex
• What happens as we change the curvature of the mirror?
IMAGE:
– Plane mirror:
virtual
» R=
upright (non-inverted)
q
q
– Convex mirror:
» R<0
h
h’
o
i
f
IMAGE:
virtual
upright (non-inverted)
Lecture 21, ACT 4
• In order for a real object to create a real, inverted enlarged image,
a) we must use a concave mirror.
b) we must use a convex mirror.
c) neither a concave nor a convex mirror can produce this image.